Given that $m \angle A= 60^\circ$, $BC=12$ units, $\overline{BD} \perp \overline{AC}$, $\overline{CE} \perp \overline{AB}$ and $m \angle DBC = 3m \angle ECB$, the length of segment $EC$ can be expressed in the form $a(\sqrt{b}+\sqrt{c})$ units where $b$ and $c$ have no perfect-square factors.  What is the value of $a+b+c$?

[asy]
draw((0,0)--(8,.7)--(2.5,5)--cycle);
draw((0,0)--(4.2,3.7));
draw((8,.7)--(1.64,3.2));
label("$B$",(0,0),W);
label("$C$",(8,.7),E);
label("$D$",(4.2,3.7),NE);
label("$E$",(1.64,3.2),NW);
label("$A$",(2.5,5),N);
[/asy]
Explanation: The diagram the problem gives is drawn very out of scale so we redraw the diagram, this time with $\overline{AC}$ as the base:

[asy]
draw((0,0)--(1+sqrt(3),0)--(1,sqrt(3))--cycle);
label("$A$",(0,0),SW); label("$C$",(1+sqrt(3),0),SE); label("$B$",(1,sqrt(3)),N);
draw((1,0)--(1,sqrt(3)));
label("$D$",(1,0),S);
draw((1+sqrt(3),0)--(.75,1.3));
label("$E$",(.75,1.3),W);
label("$y$",(2.2,.4),NW);
label("$3y$",(.95,1.55),SE); label("$60^\circ$",(.1,0),NE);
[/asy] All angles are given in degrees.

Let $\angle ECB = y$, so $\angle DBC=3y$.  From $\triangle AEC$ we have $\angle ACE = 180^\circ-60^\circ-90^\circ= 30^\circ$.

Now let $EC$ and $BD$ intersect at $F$. $\angle BFE=\angle DFC$ by vertical angles and $\angle BEF=\angle CDF=90^\circ$, so $\angle FBE=\angle FCD$, which is equal to 30 degrees.  Now summing the angles in $\triangle ABC$, we have $60^\circ+30^\circ+3y+y+30^\circ=180$, solving yields $4y=60$ so $y=15$ and we see $\triangle BDC$ is a 45-45-90 triangle.  Also, $\triangle ABD$ is a 30-60-90 triangle.

Let $ AD = x$, so $AB = 2x$ and $DB = DC = x\sqrt{3}$. $BC = x\sqrt{3}\sqrt{2} = x\sqrt{6}$.  We are given that this equals 12, so we find $x = 12/\sqrt{6} = 2\sqrt{6}$.  It follows that the area of $\triangle ABC$ can be found via  \[(1/2)(AC)(BD)=(1/2)(x+x\sqrt{3})(x\sqrt{3})=12\sqrt{3}+36.\] To find $EC$, notice that the area of $\triangle ABC$ can also be written as $(1/2)(AB)(EC)$.  Thus, \[(1/2)(4\sqrt{6})(EC)=12\sqrt{3}+36 \Rightarrow EC = 3(\sqrt{2}+\sqrt{6}).\]  Hence $a=3$, $b=2$, and $c=6$, so $a+b+c=\boxed{11}$.